Prove log(n)/n^sigma is a null sequence.

  • #1
19
0
Mod note: Moved from the homework section.
1. Homework Statement

If σ > 0 and base b > 1 then prove that ##log n/n^σ## is a null sequence. This is not really a homework since i am self studying Konrad Knopp book about infinite series and i wanted to see different ideas and perspectives on the proof of the problem.Thanks.
 
Last edited by a moderator:

Answers and Replies

  • #2
12,388
6,148
Can you provide your proof of the problem?

Also what is the b value? Is it the base of the log function?
 
  • #3
19
0
Yes the b is the base of the log function.
 
  • #4
19
0
Mod note: Edited text below to make it more readable.
If b > 1 then we have ##b^σ > 1##. Therefore ##(n/(b^σ)^n)## is a null sequence.
Given ε > 0 we have consequently from a point onwards - say for every n > m, ##(n/(b^σ)^n) < ε' = ε/b^σ##. But in any case, ##log n/n^σ < b^σ(g + 1/(b^σ)^g b^1 ))## . If g denotes the characteristic of log n . If therefore, ## n > b^m## Then, ## |(log\ n/n^σ)| < ε ∀\ n > b^m## Q.E.D

This is the proof of the book and it is a theorem not in the exercises as the author is building up the knowledge of the reader, but as i read this again i have one question now how did he come up with what n has to be bigger than ?
 
Last edited by a moderator:

Related Threads on Prove log(n)/n^sigma is a null sequence.

  • Last Post
Replies
2
Views
4K
Replies
5
Views
797
Replies
3
Views
5K
Replies
1
Views
1K
Replies
10
Views
2K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
2K
Top